Question

Show that the thickness of the ice on a lake increases with the square root of the time in cold weather, making the following simplifying assumptions. Let the water temperature be a constant \(10^{\circ} \mathrm{C},\) the air temperature a constant \(-10^{\circ},\) and assume that at any given time the ice forms a slab of uniform thickness \(x\). The rate of formation of ice is proportional to the rate at which heat is transferred from the water to the air. Let \(t=0\) when \(x=0.\)

Short answer

The thickness of the ice \( x = \sqrt{2kt} \), showing it increases with the square root of time.

Step-by-step solution

Understand the Heat Transfer Process

Heat is transferred from the water (constant at 10°C) to the air (constant at -10°C) through the ice slab. The rate at which ice forms is proportional to this heat transfer.

Define the Variables

Let the thickness of the ice be denoted by the variable \( x \) and time be denoted by \( t \). Initially, at time \( t = 0 \), the thickness \( x = 0 \).

Proportional Heat Transfer

Understand that the rate of ice formation, \( \frac{dx}{dt} \), is proportional to the rate of heat transfer from the water to the air. The heat transfer through the ice slab can be described by Fourier's law as \( Q = kA \frac{dT}{dx} \) where \( k \) is the thermal conductivity, \( A \) is the area, \( dT \) is the temperature difference between water and air, and \( dx \) is the thickness of the ice.

Relate Heat Transfer to Ice Formation

Since the heat transfer \( Q \) leads to ice formation, we can write that \( \frac{dx}{dt} \propto \frac{dT}{dx} \). With \( dT = 20 \) (the temperature difference), and \( dx = x \), this translates to \( \frac{dx}{dt} \propto \frac{20}{x} \). Introducing a constant of proportionality, \( k \), we get \( \frac{dx}{dt} = \frac{k}{x} \).

Solve the Differential Equation

Rearrange and integrate the equation: \( x \frac{dx}{dt} = k \), leading to \( x dx = k dt \). Integrating both sides, \( \int_{0}^{x} x dx = \int_{0}^{t} k dt \), we get \( \frac{x^2}{2} = kt \). Solving for \( x \), we find \( x = \sqrt{2kt} \).

Conclusion

The thickness of the ice increases with the square root of time as given by \( x = \sqrt{2kt} \).

Key concepts

Fourier's law

In physical sciences, Fourier's law describes the conduction of heat. It states that the rate of heat transfer through a material is proportional to the negative gradient in temperature and the area through which this heat is being transferred.
For a simple understanding: if you have a hot object touching a colder object, heat will flow from the hot to the cold object. Fourier’s law quantifies how fast this heat transfer will occur.
It is mathematically expressed as:

\[ Q = -kA \frac{dT}{dx} \]

Where:

• \( Q \) is the heat transfer per unit time
  \( k \) is the thermal conductivity
  \( A \) is the cross-sectional area
  \( \frac{dT}{dx} \) is the temperature gradient

This law is a cornerstone in understanding various thermal processes in engineering and natural sciences, including ice formation as in our lake example.

Differential equations

Differential equations play a crucial role in modeling and solving problems regarding rates of change in physical systems, such as heat transfer or ice formation.
They are equations that relate a function with its derivatives, meaning they capture relationships involving rates of change.
In our exercise, we consider:
\[ \frac{dx}{dt} = \frac{k}{x} \]
This equation relates the rate of change of ice thickness \( x \) with time \( t \) and involves a constant \( k \).
Rearranging and solving these equations helps us understand how the thickness of ice changes over time.

Thermal conductivity

Thermal conductivity is a material property that indicates how well a material can conduct heat.
High thermal conductivity means the material easily allows heat to pass through, while low thermal conductivity means it does not.
It is represented by the constant \( k \) in Fourier's law.
In our ice formation scenario, the rate of heat transfer from the warm water to the cold air is affected by the thermal conductivity of ice.
This property thus influences how quickly the ice forms and grows in thickness over time.

Ice formation

Ice formation on a lake during cold weather is an interesting example of heat transfer in action.
As the warm water at 10°C loses heat to the cold air at -10°C, ice begins to form at the interface and grows in thickness.
The rate of ice formation depends on the thermal properties of the ice itself and the temperature difference between the water and air.
Mathematically, we find that the thickness of the ice \( x \) increases with the square root of time \( t \), described as:

\[ x = \sqrt{2kt} \]

Proportional relationships

Proportional relationships are fundamental in describing how different quantities are related in both mathematics and physical sciences.
In our problem, the rate at which ice forms is proportional to the rate of heat transfer. This means if you increase the heat transfer rate, the ice will form faster.
We used a constant of proportionality to relate the rate of ice formation \( \frac{dx}{dt} \) to the rate of heat transfer:

\[ \frac{dx}{dt} \propto \frac{20}{x} \]
This equation implies that as the thickness of the ice increases, the rate at which it forms slows down, ensuring that this proportional relationship provides an accurate description of the physical process over time.